About corollary of sylow $p$-subgroup of finite group. I am facing a problem in group theory question or better say a doubt. If anyone can help me then please do try to solve this question. A couple of days ago, we were discussing a corollary of Sylow’s theorem in group theory (Abstract algebra) lectures. In that corollary, our teacher proved that if a finite group G has order 30 then,


*

*It will have a normal subgroup of order $15$;

*Sylow $3$-subgroup and Sylow $5$-subgroup will be normal in $G$.


If you’ll see that we can write the order of group as,
$$
  o(G) = 30 = 2 \cdot 3 \cdot 5
$$
and you can easily observe that $2,3,5$ are primes and $2 < 3 < 5$ also the difference of the first two primes is $1$ (i.e. $3-2$) and the last two primes are $2$ (i.e. $5-3$). Let us write this in quite an easy manner,
$$
 (2, 3, 5) ⇒ (1, 2)
$$
where the product of primes in left side gives us the order of G and the right side’s no. are the differences between the first two and the last two primes respectively. Though I know that this is not the formal way of writing this, for the sake of convenience, I’m writing it in this way. 
Now my doubt is, are there infinitely many such combinations of three primes $(p, q, r)$ with $p < q < r$ such that the difference of first two and the last two primes can be written as $2^{k-1}$ (i.e. $q-p$) and $2^k$ (i.e. $r-q$), where $k \in \mathbb{N}$, i.e.
$$
 (p, q, r) ⇒ (2^{k-1}, 2^k)
$$
such that $p \cdot q \cdot r$ is an order of a group $G$ such that,


*

*$G$ has a normal subgroup of order $q \cdot r$;

*Sylow $q$-subgroup and Sylow $r$-subgroup will be normal in $G$.


Moreover, if we are able to find such combination $(p, q, r)$ for some $k \in \mathbb{N}$, then will it be a unique combination for that $k$ and if it is not unique i.e. if there is more than one such combination then how many combinations are possible for that particular $k$ and is there any relation between the combinations for that $k$?
 A: Setting the record straight from a group-theoretic perspective: the remark of Steve is of course very valid: if you require all the Sylow subgroups to be normal, then $G$ is nilpotent, and since their order are $p$, $q$, and $r$ respectively, we get $G \cong C_{pqr}$, which is kind of boring. But, irrespective of the differences between the choices of the primes $p$, $q$ and $r$, the following is true.
Theorem Let $|G|=pqr$, with $p \lt q \lt r$ prime. Then the Sylow $r$-subgroup of $G$ is normal.
To prove this we need a lemma for two primes.
Lemma Let $|G|=pq$, with $p \lt q$ prime. Then the Sylow $q$-subgroup of $G$ is normal.
Proof Of course $n_q \in \{1,p\}$. If $n_q=p$, then $p \equiv 1$ mod $q$, implying $q \leq p-1 \lt p$, contradicting $p \lt q$. Hence $n_q=1$.

Before diving into the proof of the theorem, let us digest on what this implies. So, let $|G|=pqr$, with primes $p \lt q \lt r$. According to the theorem, there is a unique and hence normal Sylow $r$-subgroup, say $R$. The lemma tells us that $G/R$ has a unique Sylow $q$-subgroup, being $QR/R$, where $Q \in Syl_q(G)$ (I am using here that all Sylow subgroups of factor groups arise as quotients of their corresponding Sylow subgroups of the whole group). We conclude:
Corollary Let $|G|=pqr$, with $p \lt q \lt r$ prime and let $Q \in Syl_q(G)$ and $R \in Syl_r(G)$. Then $QR$ is normal in $G$. 
Now let us prove the theorem.
Observe that $n_r \in \{1,p,q,pq\}$ and assume that $n_r \neq 1$. We will derive a contradiction. First, we can rule out $n_r=p$, since this would imply $p \equiv 1$ mod $r$, whence $r \leq p-1$, contradicting $p \lt r$. Similarly, $n_r \neq q$. Hence $n_r=pq$. So, the number of elements of order $r$ equals $pq(r-1)=|G|-pq$. We need this later.
Now let us concentrate on $n_q$. Of course $n_q \in \{1, p, r, pr \}$, but $p \lt q$ implies that $n_q \neq p$. We will argue that $n_q=1$: assume for the moment that $n_q \geq r$, then the number of elements of order $q$ is at least $r(q-1)$. Combining this with what we know about the number of elements of order $r$ we get $|G| \geq r(q-1)+|G|-pq$, so $pq \geq r(q-1)$. But $pq \lt pr$, amounting to $q-1 \lt p$, which is nonsense. So we must have $n_q \lt r$, leading to $n_q=1$ as the only possibility.  Now we apply the lemma (twice!): let $Q$ be the unique Sylow $q$-subgroup of $G$ and let $R$ be a Sylow $r$-subgroup of $G$. We have shown $Q \lhd G$ and the lemma yields $QR/Q \lhd G/Q$, so $QR \lhd G$. Of course $R$ is Sylow in $QR$ and again by the lemma, even $R \lhd QR$. Now we have $R$ char $QR \lhd G$, so finally $R \lhd G$, contradicting $n_r=pq$. $\square$
Remarks You could ponder on what happens if $|G|=pqrs$, with $p \lt q \lt r \lt s$, primes (one can show that if $|G|$ is square-free then the Sylow $p$-subgroup is normal, where $p$ is the largest prime dividing $|G|$). Also, in the theorem and the corollary, one can prove that $Q \lhd G$ if and only if $q \nmid r-1$.
